Let $f(x) = 2x^4-17x^3+26x^2-24x-60$. Find $f(7)$.
Solution: Instead of plugging in $x=7$ into $f(x)$ and solving, we can use the Remainder Theorem to avoid complicated arithmetic. We know that $f(7)$ will be the remainder when $f(x)$ is divided by $x-7$. So we have:

\[
\begin{array}{c|ccccc}
\multicolumn{2}{r}{2x^3} & -3x^2&+5x&+11 \\
\cline{2-6}
x-7 & 2x^4 &- 17x^3 &+ 26x^2&-24x&-60  \\
\multicolumn{2}{r}{2x^4} & -14x^3  \\ 
\cline{2-3}
\multicolumn{2}{r}{0} & -3x^3 & +26x^2   \\
\multicolumn{2}{r}{} &-3x^3  &+21x^2   \\ 
\cline{3-4}
\multicolumn{2}{r}{} & 0& 5x^2 & -24x   \\
\multicolumn{2}{r}{} & & 5x^2 & -35x   \\
\cline{4-5}
\multicolumn{2}{r}{} & & 0 & 11x & -60   \\
\multicolumn{2}{r}{} & &  & 11x & -77   \\
\cline{5-6}
\multicolumn{2}{r}{} & &  & 0 & 17  \\
\end{array}
\]Hence $f(7) = \boxed{17}$.